Question

Show that the lines $\overline{r}=\left(\stackrel{\wedge }{i}+\stackrel{\wedge }{j}-\stackrel{\wedge }{k}\right)+\lambda \left(3\stackrel{\wedge }{i}-\stackrel{\wedge }{j}\right)$ and $\overline{r}=\left(\stackrel{\wedge }{4i}-\stackrel{\wedge }{k}\right)+\mu \left(2\stackrel{\wedge }{i}-3\stackrel{\wedge }{k}\right)$ intersect. Find their point of intersection.

Answer 1

Here, we have, $r_1=i+j-k+\lambda (3i-j)$ and $r_2=4i-k+\mu (2i-3k)$

To show that $r_1$ and $r_2$ intersects we must first equalize them

$\therefore$ $r_1=r_2$

$\therefore$ $i+j-k+\lambda (3i-j)=4i-k+\mu (2i-3k)$

$\therefore$ $(3\lambda -2\mu -3)i+(1-\lambda )j+k\left(3\mu \right)=0$

Now, equating the $i,j,k$ each component to zero we get,

First equating $k$ component as zero, we get,

$-3\mu =0$

$\therefore \mu =0$

Then equating $j$ component as zero, we get,

$1-\lambda =0$

$\therefore \lambda =1$

Lastly equating $i$ component as zero, we get,

$3\lambda -2\mu -3=0$

$\therefore 3-3=0$

Hence, $r_1$ and $r_2$ intersects each other.

$\therefore$ Intersecting point : $\mu =0,\lambda =1$

Also, when we put the values of $\lambda$ and $\mu$ in any of the above equation we will get the point of intersection as: $4i-k.$